# Construction a multiplicative inverse of a real number via Dedekind cuts

I'm reading Pugh's "Real Mathematical Analysis" where C.Pugh constructs real numbers system using Dedekind cuts. Unfortunately, he omits a construction of a multiplicative inverse of a real number $$x = A|B$$ where $$A$$ and $$B$$ are cuts in $$\mathbb{Q}$$. First, some required definitions:

A cut in $$\mathbb{Q}$$ is a pair of subsets $$A,B$$ of $$\mathbb{R}$$ such that

a)$$A \cup B = \mathbb{Q}, A \neq \varnothing, B \neq \varnothing, A \cap B = \varnothing$$

b) If $$a \in A$$ and $$b \in B$$ then $$a < b$$

c) A contains no largest element

A real number is a cut in $$\mathbb{Q}$$.

The additive inverse of a cut $$x = A|B$$ is defined as $$(-x) = C|D$$ where $$C = \{r \in Q| \exists b \in B$$ (not the smallest elements of $$B$$ ) $$: r = -b \}$$ and $$D = \mathbb{Q} \setminus C$$

Multiplication is defined for positive numbers $$x > 0$$. If $$x = A|B, y = C|D$$, then $$xy = E|F$$ where

$$E = \{r \in \mathbb{Q}| r \leq 0$$ or $$\exists a \in A \ \ \exists c \in C: a > 0, c > 0, r = ac \}, F = \mathbb{Q} \setminus E$$. If $$x > 0, y < 0$$ then we define $$xy = -(x(-y))$$

Now I wonder how can we define a multiplicative inverse of a positive cut $$x = A|B$$. Obviously, it's a cut $$\frac{1}{x} = C|D$$ such that $$x \frac{1}{x} = 1$$

So, it's such a cut $$C|D$$ such that for $$C$$ we have $$\{r \in \mathbb{Q}| r \leq 0$$ or $$\exists a \in A \ \ \exists c \in C : a > 0, b > 0, r = ac \} = \{r \in \mathbb{Q}| r < 1 \}$$

Any ideas how we can define $$C$$(if we define $$C$$, then $$D = \mathbb{Q} \setminus C$$)?

Edit: Changed to correct the error modnar points out in the comments.

If $$x > 0$$, $$C = \{r \in \Bbb Q \mid \exists b \in B,\, rb < 1\}$$ If $$x < 0$$, $$C = \{r \in \Bbb Q \mid \exists b \in B,\, b < 0\text{ and } rb > 1\}$$

Original post: (has maximums when $$x$$ is rational)

If $$x > 0$$, $$C = \{ r \in \Bbb Q \mid r \le 0 \text{ or } \forall a \in A,\, ra < 1\}$$

If $$x < 0$$, $$C = \{ r \in \Bbb Q \mid \forall a \in A, \,ra > 1\}$$

• When $a$ is the multiplicative identity in $\mathbb{R}$, i.e, $a=\{r\in \mathbb{Q}\mid r<1\}$, the above $C$ becomes $a\cup \{1\}$, which has a greatest element $1$. – modnar Apr 11 at 18:23
• @modnar - it took a while to interpret your comment, because the first time you mentioned "$a$", you actually meant "$x$", and the other two times you mentioned "$a$", you actually meant "$A$". However, you are correct, and the problem is not just with $1$, but occurs any time $x$ is rational. I've changed the answer to one version that doesn't have maximums. A different fix would have been to just add $r \ne \min B$ to each of the original set definitions, but then I'd have to point out this still works when $\min B$ doesn't exist. – Paul Sinclair Apr 11 at 21:33
• You are right. The notation in my comment is confusing. I like your revised construction, which is quite elegant. – modnar Apr 12 at 10:31
• I am confused. Why did you not simply take $C=\{\frac{1}{y}:x\ne y\in B\}$? – Shahab Apr 25 at 8:29
• @Shahab - when $x < 0$, your $C$ will contain both positive elements that are $> \frac 1x$ and negative elements that are $< \frac 1x$. When $x > 0$, your $C$ will not contain $0$ itself, which the full lower Dedekind cut should include. And even for non-zero rationals, my version obviously contains every rational $<\frac 1x$, while yours requires an additional (though simple) argument to show it. And another reason is that my version doesn't require the development of division in $\Bbb Q$ before it can be introduced. – Paul Sinclair Apr 25 at 13:31

A possible way to fix the bug mentioned under Paul's answer is $$C=\{r\in \mathbb{Q}\mid r\leq 0\ \text{or}\ (\exists s\in \mathbb{Q}\ (r for the case $$x>0$$. A similar trick should be able to take care of the case $$x<0$$, too.

• If you properly qualify $a$, that will work. As is, $C = \emptyset$, since for any rational $s$, there are objects $a$ for which multiplication with $s$ is defined and $sa > 1$. – Paul Sinclair Apr 11 at 21:40
• @Paul, Indeed. $\forall a$ should be $\forall a\in A$. Thanks. – modnar Apr 12 at 10:06